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$R=\{a+bi|a,b\in \mathbb{Z}\}$, Prove $R/(1+i)$ is a field.

I can write the element of $R/(1+i)$ exactly. Actually $R/(1+i)=\{\bar{0},\bar{1}\}$. And I can examine every condition that make a set to be a field. But I think this makes it a little inelegant. What's the more effective ways.

Jaqen Chou
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  • Show that $(1+i)$ is a maximal ideal - which follows from the fact that there are less than four cosets. Btw, are you aware that $(1-i)(1+i)=2$? – Hagen von Eitzen Oct 03 '18 at 11:50
  • Taking a field modulo a nonzero field element leads you nowhere. – Wuestenfux Oct 03 '18 at 12:02
  • You overcounted the elements of $R/(1+i)$. We have $-1=1$ in $R/(1+i)$. Note that $1-(-1)=2=(1-i)(1+i)=0$ as an element of $R/(1+i)$. –  Oct 03 '18 at 12:14

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We have $$ R/(1+i) \cong \mathbb Z[x]/(x^2+1,1+x) = \mathbb Z[x]/(1+x,2) \cong \mathbb Z/(2) \cong \mathbb F_2 $$ Indeed, $2=(x^2+1)+(1+x)(1-x)$ implies $$ (x^2+1,1+x) =(x^2+1,1+x,2) =(x^2-1,1+x,2) =(1+x,2) $$

lhf
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